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V 1 L L E L : S Cartcr (inily < 4xintet 1 N U V 1 Katet (itnut M MSu L The...

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V 1 L L E L : S Cartcr (inily < 4xintet 1 N U V 1 Katet (itnut M MSu L The

V 1 L L E L : S Cartcr (inily < 4xintet 1 N U V 1 Katet (itnut M MSu L The



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$\mathrm{K}_{2} \mathrm{CrO}_{1}$ givcs a ycllow precipitare on rcaction with: (a) $C_{4}^{2}$ (b) $\mathrm{Fe}^{3}$ (c) $\mathrm{Ba}^{2}$ (d) $\mathrm{Pb}^{2}$

And a the negative stays out front in the absolute value of negative on his positive one. So negative one is our answer, and B the two negatives make a positive, so positive one and see the absolute value of one is one D. The absolutely of negative one is one in E. The negative stays out front in the AB. Survive one is once or answer is negative one.

They're. So for this exercise we have this vector B. And the subspace dovey generated by the one, V two and V three that are these vectors that are defined here. So basically we need to calculate the Earth a little projection of you on this space to view. And just remember remember this projection is calculated as the inner proud of the vector V. Each of the generators of this subspace dog. In this case the generators RV one, The two and 3. So we need to calculate the we need to calculate the inner part of me with each of the generator divided the score of the norm of the generators times degenerates. So these for the three vectors B two square plus the interpreter of B would be three. B three. Did the square of the norm of B. Three. Okay, so just to remind you a little bit of the geometric intuition of this, is that the view is generated by these three vectors. So what we're doing is projecting we on each of the generators and then some that together. So we want We t. v. one and V three acts as a basis. Actually in this case they are linearly independent so they form a basis for this. Yeah, subspace of you. So we're writing the in terms of this basis. So we're projecting projecting on this sub space. So let's calculate the correspondent values that we need. So in this case we would be one. The product of B would be to dinner product of the would be three. So this is equal two, one half, There is a constitute and this inner product is equal to zero and then the norms. So because this is the cost to zero means that we don't need this term anymore is going to be equal to zero. So we just need to calculate the score of the norms for B. two and B one. So for me, one square of the norm, remember that there is equal to the inner product of the vector with itself. And in this case this result in one and the inner approach of B two square is equal 2, 1 as well. So these are actually military vectors. And then we just need to put all together on the four. So behalf that the projection of the vector B on the subspace, our view, it's equals to 1/4 times 11 one plus the vector V two. That is equal to one, 1 -1 -1. After some. In these two vectors obtain the action solution that is one half times the vector, three, three minus one minus one. That corresponds to their thermal projection of beyond this subspace of you.

In the problem we have 56. And this is yeah it is gen X get works. Get three X. Get your necks. It's two weeks is tricks plus if one X after works three X gen X G networks. GTX H one X. HdX history X plus. If you're next afterwards after the X. And the one x. Networks G three X H one X H two x. and three x. No further. We have fds a. And this equals two. If one yeah. Have to wear after a year. Do you want to G two A. Gayatri? H one A. It's too good. Is three. Okay. And these are reputed To two times more studies if one A. Have to A F three given a and geeta, Geeta A 20 H two A. It was today glass if one after where have today do you want to? They were the tree H one A H two A is three. Now this equals to zero plus zero plus zero which is equal to zero. Since F R A Is equal to zero. G R A is equal to H R A. Is I mean this equals to 04. All the times that is if R0 gr hr is also zero For Article 1, 2 and three. So we have this as the answer to the problem

In this video, we're gonna go through the answer to question number 17 from chapter 9.4. So we asked to find where these vectors x one x two x three Ah, linearly dependent on where they are linearly independence, which FYI is teeth independent much varsity that linearly dependent. So okay, so for them to be linearly dependent would need values off. See one c two c three it such that c one times x one c two times x two plus C three times Next three equal to zero association into values for these x one x two x three then we write This is a system of three equations. So the 1st 1 is just gonna be Ah, well, it's the the first element of each of the equations Time each of the vectors X one experience to text three times by the corresponding um constant C one C to C three. So we're gonna have Well, there's no, uh, ex threes are no component, Maxime, because the top of the next three is zero. So we're going off, uh, ex one next Tuesday, at both of which contain eats the to tease it. So take a common factor out. Get either to t C one plus C two. Ah equals there. Next up, we're gonna have eats the two tea. Close. It was C one that's actually to eat the TT plus C three he to the three tea equal Syria. And finally five. Easter to tee times five C one minus C two equals Sarah. Okay, so comparing the first of these equations on the last of these equations we find that C one don't see too must be equal to zero as you see that because, well, first look in the first equation e to the two tea for any tea can never be zero. So we basically just cancel by its beauty and saying with the bottom equation s So therefore, we have this bit is equal to zero this busy zero for those two to both equal to zero. Then C one and C t o. You can show that by rearranging one of those sub student in the other. Um and you're find this. You wanna see t birthday frequent zero. So therefore I see two is the rocks that this is zero. So then we have to see three times Three tier sequence There we can we can divide by eats and three team because that's never zero for any tea on DDE. All we're left with is C three sequences So far, all t ah the C one c to the T three. Must could only be for this for this equation. Thio be satisfied. Uh, this equation to be satisfied then Theo Dissolution of the Trivial Solution. Therefore the vectors x one x two x three are linearly independent for that tea in any value between minus infinity to infinity.


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